Question: $f(x, y, z) = (\tan(y), z\sin(x), 2\cos(x))$ $\text{curl}(f) = $ $\hat{\imath} + $ $\hat{\jmath} + $ $\hat{k}$
Solution: $f(x, y, z) = (f_0, f_1, f_2)$ The curl of $f$ : $\begin{aligned} \text{curl}(f) &= \det \begin{bmatrix} {\hat{\imath}} & \hat{\jmath} & \hat{k} \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ f_0 & f_1 & f_2 \end{bmatrix} \\ \\ &= \left( \dfrac{\partial f_2}{\partial y} - \dfrac{\partial f_1}{\partial z} \right) \hat{\imath} \\ \\ &+ \left( \dfrac{\partial f_0}{\partial z} - \dfrac{\partial f_2}{\partial x} \right) \hat{\jmath} \\ \\ &+ \left( \dfrac{\partial f_1}{\partial x} - \dfrac{\partial f_0}{\partial y} \right) \hat{k} \end{aligned}$ $\begin{aligned} f_0(x, y, z) &= \tan(y) \\ \\ f_1(x, y, z) &= z\sin(x) \\ \\ f_2(x, y, z) &= 2\cos(x) \end{aligned}$ Let's calculate all the partial derivatives we'll need. $f_0$ $f_1$ $f_2$ $\dfrac{\partial}{\partial x}$ $z\cos(x)$ $-2\sin(x)$ $\dfrac{\partial}{\partial y}$ $\sec^2(y)$ $0$ $\dfrac{\partial}{\partial z}$ $0$ $\sin(x)$ Now we can put it all together. $\begin{aligned} \text{curl}(f) &= \left( \dfrac{\partial f_2}{\partial y} - \dfrac{\partial f_1}{\partial z} \right) \hat{\imath} \\ \\ &+ \left( \dfrac{\partial f_0}{\partial z} - \dfrac{\partial f_2}{\partial x} \right) \hat{\jmath} \\ \\ &+ \left( \dfrac{\partial f_1}{\partial x} - \dfrac{\partial f_0}{\partial y} \right) \hat{k} \\ \\ &= (0 - \sin(x)) \hat{\imath} + (0 + 2\sin(x)) \hat{\jmath} \\ \\ &+ (z\cos(x) - \sec^2(y)) \hat{k} \\ \\ &= -\sin(x) \hat{\imath} + 2\sin(x) \hat{\jmath} + (z\cos(x) - \sec^2(y)) \hat{k} \end{aligned}$ We could also write the $\hat{k}$ -component as $z\cos(x) - \dfrac{1}{\cos^2(y)}$ because $\sec(y) = \dfrac{1}{\cos(y)}$. In conclusion: $\text{curl}(f) = -\sin(x) \hat{\imath} + 2\sin(x) \hat{\jmath} + (z\cos(x) - \sec^2(y)) \hat{k}$